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Google·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say
Apr 2026

Summary

Google coding interview, one tree problem, nothing too fancy but it made me think harder than I expected.

Questions Asked (1)

Q1

Given two binary trees, write a function to determine whether they are identical.

Algorithms & Data Structures
Author's notes

Went straight to recursion, which felt right.

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AI HintsAI Generated

Suggested Approach

Clarify the definition of 'identical' (structure and node values) and then present a recursive solution that compares the roots and recursively checks left and right subtrees. Discuss iterative alternatives (e.g., using a stack or queue) and analyze time and space complexity.

Pro tip: Mention that the recursive solution can cause stack overflow for very deep trees, so an iterative approach using an explicit stack might be preferable in production systems. Also, note that early termination on mismatch makes the average case faster.

1. Clarify the problem

Confirm that 'identical' means both structure and node values must match exactly. Ask about edge cases like empty trees or trees with one node.

2. Choose an approach

Decide between recursive and iterative solutions. Recursive is simpler and more elegant; iterative avoids stack overflow for deep trees.

3. Outline the algorithm

For recursion: if both nodes are null, return true; if one is null, return false; if values differ, return false; otherwise recurse on left and right subtrees. For iteration: use a stack (or queue) to simulate recursion.

4. Analyze complexity

Time complexity is O(n) where n is the number of nodes (or min(n,m) if sizes differ). Space complexity is O(h) for recursion (h = height) or O(n) for iterative in worst case.

5. Test with examples

Walk through simple cases: both empty, one empty, same values, different values, different structures. Also consider large trees and skewed trees.

Key Points to Mention

  • Definition of identical: same structure and same node values.
  • Recursive solution: base cases for null nodes, value comparison, and recursive calls.
  • Iterative solution using stack or queue to avoid recursion depth issues.
  • Time complexity O(n) and space complexity O(h) for recursion, O(n) for iterative.
  • Early termination when a mismatch is found.
  • Edge cases: empty trees, single node, skewed trees, duplicate values.

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.